id stringlengths 5 5 | fame_level stringclasses 3
values | fame_level_original stringclasses 2
values | domain stringclasses 6
values | case int64 1 6 ⌀ | source stringlengths 34 76 | source_licence stringclasses 2
values | verbatim_status stringclasses 7
values | gold_status stringclasses 2
values | split stringclasses 1
value | problem_statement stringlengths 48 179 | gold_answer stringlengths 5 132 | gold_method stringlengths 121 493 | web_frequency stringclasses 5
values | web_frequency_note stringlengths 65 523 | expected_contamination stringclasses 4
values | languages_available_in_release listlengths 1 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
F1-01 | F1 | null | algebra | 4 | Rosen 1831, p.8 (Kitab al-jabr, type ‘mal wa jidhr’) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | What must be the square which, when increased by ten of its own roots, amounts to thirty-nine? (x² + 10x = 39) | x = 3 (square = 9) | Case 4, mal wa jidhr ya'dilu adad. Halve the roots (5); square it (25); add to 39 → 64; root is 8; subtract half the roots (5) → 3. Geometric justification: square of side x with two 5-wide rectangles on adjacent sides, completed by a 5×5 corner (gnomon), total area 64, side 8, so x = 3. | high | Verbatim on Wikipedia, Britannica, MacTutor, Harvard, dozens of tutorials. Maximal contamination. Control item. | very_high | [
"en"
] |
F1-02 | F1 | null | algebra | 5 | Rosen 1831, pp.16-18 (type ‘mal wa adad ya'dilu jidhr’) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | A square and twenty-one dirhems are equal to ten roots. What is the root? (x² + 21 = 10x) | x = 3 or x = 7 | Case 5. Halve the roots (5); square it (25); subtract the 21 → 4; root is 2; add to and subtract from half the roots → 5±2 = 7 or 3. Al-Khwarizmi notes the two-root structure and the non-solvability caveat when (b/2)² < c. Geometric demonstration via Euclid II.5-type figure. | high | On MacTutor, Berggren ch.3 (online PDF), Britannica context. High contamination. | high | [
"en"
] |
F1-03 | F1 | null | algebra | 1 | Rosen 1831 (type ‘mal ya'dilu jidhr’) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | A square is equal to five of its roots. What is the root? (x² = 5x) | x = 5 (square = 25) | Case 1, mal = jidhr. The root equals the number of roots. Al-Khwarizmi does not treat x = 0 as a solution (zero is not admitted as a root). No completing the square needed. | med | Rosen 1831 (PD, archive.org OCR): 'A square is equal to five roots of the same.' Six-case taxonomy ubiquitous; instance coefficient confirmed. med. | high | [
"en"
] |
F1-04 | F1 | null | geometry | null | Rosen 1831, mensuration section (area of the circle) | public domain (Rosen 1831) | method_verbatim_instance_applied | verified | public | What is the area of a circle whose diameter is ten? | ≈ 78 4/7 (using π ≈ 22/7) | Al-Khwarizmi's rule: multiply the diameter by itself, then subtract one-seventh and half of one-seventh (i.e. multiply d² by 1 − 1/7 − 1/14, equivalent to π ≈ 22/7 over d²/4 expressed his way). 10×10 = 100; 100 − 100/7 − 100/14 ≈ 78 4/7. He also mentions the geometers' ‘more exact’ rule and the astronomers' 62832/20000... | med | Rosen gives the circle-area RULE verbatim (multiply diameter by itself, subtract 1/7 and half 1/7 = d²×11/14); d=10 worked case is a standard application, answer 78 4/7 math-verified. | med | [
"en"
] |
F1-05 | F1 | null | inheritance | null | Rosen 1831, legacies chapter (archive.org OCR ~p.65) | public domain (Rosen 1831) | statement_verbatim_answer_TBD | candidate_pending_expert_check | public | A man dies, leaving two sons behind him, and bequeathing one-third of his capital to a stranger. He leaves ten dirhems of property and a claim of ten dirhems upon one of the sons. | CANDIDATE: amount recovered from the indebted son ("thing") = 5 dirhems (stranger takes 1/3). Verify split vs Rosen worked solution. | Al-Khwarizmi normalizes on the legal share, treats the unknown share as a thing (shay'), and solves by algebra subject to the rule that a bequest to a stranger may not exceed one-third without the heirs' consent. Reproduce his share-based reasoning, not a modern system of linear equations. | med | The bequest chapter is summarized in histories but specific instances are less reproduced. Good F1/F2 boundary item once fixed. | med | [
"en"
] |
F2-01 | F2 | null | algebra | 6 | Rosen 1831 (type ‘jidhr wa adad ya'dilu mal’) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | Three roots and four of simple numbers are equal to a square. What is the root? (3x + 4 = x²) | x = 4 (square = 16) | Case 6, jidhr wa adad = mal. Halve the roots (1.5); square it (2.25); add the 4 → 6.25; root is 2.5; add half the roots (1.5) → 4. Geometric justification analogous to case 4/5. | med | Sixth case is named in every taxonomy list; the instance 3x+4=x² appears in some slides (e.g. confirmed in search). Confirm Rosen's own numbers; if different, mark as constructed-in-method. | med | [
"en"
] |
F2-02 | F2 | null | algebra | 2 | Rosen 1831 (type ‘mal ya'dilu adad’) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | A square is equal to nine. What is the root? (x² = 9) | x = 3 (square = 9) | Case 2, mal = adad. Direct extraction of the root; al-Khwarizmi treats the mal as an area whose side is sought. No negative root admitted. | med | Trivial instance; case named everywhere but specific number rarely discussed alone. | med | [
"en"
] |
F2-03 | F2 | null | algebra | 3 | Rosen 1831 (type ‘jidhr ya'dilu adad’) | public domain (Rosen 1831) | frame_not_verbatim | verified | public | Five roots are equal to ten. What is the root? (5x = 10) | x = 2 | Case 3, jidhr = adad. Divide the number by the number of roots. Linear; al-Khwarizmi treats it as the simplest reduction. | med | Rosen's roots=number instances differ ('one root is equal to five', 'half the root equals ten'); '5x=10' is representative. Lock exact instance before real run. | med | [
"en"
] |
F2-04 | F2 | null | geometry | null | Rosen 1831, mensuration chapter (archive.org OCR ~p.79) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | An equilateral triangle with acute angles, every side of which is ten yards long. Find its area. | ≈ 43.30 (= 25√3; height = √75) | Drop the perpendicular from the apex to the base; by the right-triangle relation the height is 8 (since 10² − 6² = 64). Area = half the base times the height = 6 × 8 = 48. Al-Khwarizmi gives a geometric (area-decomposition) justification rather than a formula. | low | Replaces the earlier isosceles 10-10-12 (not in Rosen) with Rosen's verbatim equilateral side-10 instance; area via surd √75. | low | [
"en"
] |
F2-05 | F2 | null | commerce | null | Rosen 1831, mercantile transactions chapter (archive.org OCR ~p.68) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | Ten for six — how much for four? (rule of four quantities / mercantile proportion) | 6 2/3 (multiply 10×4=40, divide by 6) | Al-Khwarizmi's 'rule of four quantities' (proportion): of four quantities three are known and the fourth is sought; the answer is found by cross-multiplication and division (the known opposite the unknown times the third, divided by the first). Reproduce his rhetorical rule, not a modern equation. | low | Rosen's own arrangement gives 6 2/3 (the '4' is a sum of money, answer is quantity) — NOT the 2.4 'price of four' reading. | low | [
"en"
] |
F3-01 | F3 | null | algebra | 5 | Reduction problem in al-Khwarizmi / Abu Kamil tradition; cf. Rashed 2009 | item statement and gold derivation are our own formulation; the modern edition is cited, not reproduced | instance_constructed_frame_on_web | verified | public | Divide ten into two parts such that the sum of their squares is fifty-eight. What are the parts? | the two parts are 3 and 7 | Let one part be a thing (shay'), the other 10 − thing. Their squares: thing² + (10−thing)² = 58 → 2 mal + 100 − 20 jidhr = 58. By al-jabr (restore the deficient roots) and al-muqabala: 2 mal + 42 = 20 jidhr → reduce to one mal: mal + 21 = 10 jidhr (case 5). Solve → 7 and 3. Reproduce the al-jabr/al-muqabala reduction e... | med_frame_low_instance | Answer 3,7 math-verified (9+49=58); reduces to x²+21=10x. Rosen's exact statement is OCR-corrupted in the scan — confirm wording vs print. Frame HIGH on Wikipedia, instance LOW. | med | [
"en"
] |
F3-02 | F3 | null | algebra | 4 | Rosen 1831 (archive.org OCR ~p.39) | public domain (Rosen 1831) | verbatim_confirmed | verified | public | A square-root which, when multiplied by its third, amounts to ten. | root = √30 (root² = 30) | Reduce to a standard case by dividing through by the coefficient of mal (al-Khwarizmi: 'reduce to a single square'); then apply the standard rule for the resulting case. Keep the unknown as mal/jidhr, no symbolic manipulation. | low | Verbatim fractional-coefficient root problem: x·(x/3)=10 → x²=30. | low | [
"en"
] |
F3-04 | F3 | null | inheritance | null | Rosen 1831, legacies chapter (archive.org OCR ~p.66) | public domain (Rosen 1831) | statement_verbatim_answer_TBD | candidate_pending_expert_check | public | A man dies, leaving two sons and ten dirhems of capital and a demand of ten dirhems against one of the sons, and bequeaths to a stranger one-fifth of his property and one dirhem. | CANDIDATE: amount taken from the debt ("thing") = 5 5/6 dirhems. Verify vs Rosen worked solution. | Normalize on the legal share; set the unknown share as shay'; build the relation enforcing that the stranger's bequest does not exceed one-third; solve by al-jabr. Reproduce the legal-arithmetic reasoning step by step. | low | Complex bequest instances essentially absent from popular web. Strong F3. | low | [
"en"
] |
F4-01 | F2 | F2_REVISED_was_F4 | geodesy | null | al-Biruni, Tahdid nihayat al-amakin, tr. Jamil Ali 1967; Nandana, 1017 CE | item statement and gold derivation are our own formulation; the modern edition is cited, not reproduced | method_confirmed_RESTRATIFIED | verified | public | From the summit of a mountain of height h the angle of dip of the horizon is measured as alpha. Determine the radius of the Earth R by al-Biruni's method. | R ≈ 6335 km (al-Biruni's value ≈ 12,803,337 cubits) | The line of sight from the summit is tangent to the sphere, hence perpendicular to the radius at the point of tangency. In the right triangle (Earth's centre – summit – tangency point) al-Biruni uses the relation of CHORDS / SINES (law of sines applied via the period's chord tables) and sexagesimal arithmetic. The mode... | HIGH | Restratified F4 → F2. This task is on Wikipedia, Britannica, MacTutor, ThatsMaths, Owlcation, multiple blogs, Scribd, academic PDFs. Critically, the public web is SPLIT: some sources present the law-of-sines derivation, others give the bare cos-alpha formula. A model emitting 'cos alpha' is plausibly reproducing the do... | high | [
"en"
] |
F4-02 | F3 | F3_REVISED_was_F4 | trigonometry | null | al-Kashi, treatise on the chord / Miftah al-hisab; Risala al-watar wa'l-jaib | item statement and gold derivation are our own formulation; the modern edition is cited, not reproduced | method_confirmed | verified | public | How did al-Kashi compute the sine of one degree? | sin 1° ≈ 0.0174524064... (al-Kashi gave a sexagesimal value accurate to ~17 places) | Al-Kashi reduced the problem to the cubic relating sin(3θ) and sin(θ): sin 3θ = 3 sinθ − 4 sin³θ, knowing sin 3° from constructible angles, then solved the cubic by ITERATIVE successive approximation (fixed-point iteration) in the SEXAGESIMAL system. Both the iterative method and sexagesimal working are required; a mod... | med | RESTRATIFICATION SUGGESTED. The cubic-iteration story is on Wikipedia and several history pages, though less saturated than al-Biruni. Closer to F3 than F4 on an objective axis. | med | [
"en"
] |
KHWAREZM-100
An access-bounded benchmark of medieval Khwarezm-school mathematics.
KHWAREZM-100 asks a language model to solve problems from al-Khwarizmi's Kitab al-jabr wa-l-muqabala and companion Khwarezm-school sources (al-Biruni, al-Kashi), and scores two things separately:
- Correctness: is the final answer right? (rubric criterion 9, verified numerically and symbolically)
- Method fidelity: is the answer reached by the source's own method, without anachronism? (rubric criteria 1–8, rated by blind human annotators; no LLM judge)
A model that solves x² + 10x = 39 with the quadratic formula scores full correctness and zero fidelity. A model that halves the roots, squares, completes the area and extracts the root, as al-Khwarizmi did, scores both.
Release status: v0.1.0 (pilot). 15 public items of the 100 planned. This is a mirror of the Zenodo release; the evaluation harness and power-analysis script are on GitHub.
Usage
from datasets import load_dataset
ds = load_dataset("rav11l/khwarezm-100", split="test")
print(ds[0]["problem_statement"])
Give the model problem_statement only. gold_answer and gold_method are references for scoring. Fidelity must be rated by human annotators with the rubric in RUBRIC.md; see the harness on GitHub for blind, shuffled annotation sheets.
Composition
| Fame level | Items | Gold verified | Gold candidate |
|---|---|---|---|
| F1 (most reproduced on the web) | 5 | 4 | 1 |
| F2 | 6 | 6 | 0 |
| F3 | 4 | 3 | 1 |
| F4 (least reproduced) | 0 | – | – |
| Total | 15 | 13 | 2 |
Domains: algebra 8, geometry 2, inheritance 2, commerce 1, geodesy 1, trigonometry 1.
Fame levels are a measured proxy for training-data contamination, assigned by a web-frequency protocol (instance, frame and method queries), not by general reputation. Two items drafted as F4 were restratified to F2 and F3 after measurement, so the F4 stratum is empty in this release.
Fields
| Field | Meaning |
|---|---|
id |
Stable item ID; the prefix is the originally drafted fame level |
fame_level, fame_level_original |
Measured fame level F1–F4; original level if restratified |
domain, case |
Domain; al-Khwarizmi's equation type 1–6 where applicable |
source, source_licence |
Source of record and licence status of the wording |
verbatim_status |
How closely the statement follows the source |
gold_status |
verified or candidate_pending_expert_check |
split |
public for all items in v0.1.0 |
problem_statement |
The prompt given to the model |
gold_answer, gold_method |
Reference answer and reference derivation in the source's method |
web_frequency, web_frequency_note, expected_contamination |
Evidence behind the fame level |
languages_available_in_release |
["en"] in v0.1.0 |
Full definitions: DATASET_CARD.md.
Known limitations
- Pilot only: 15 of 100 items. The fame–fidelity correlation claim needs at least 33 source-locked items; this pilot reports design power, not that result.
- Two items (F1-05, F3-04) carry candidate gold answers; exclude them from correctness scoring until verified.
- English statements only. Arabic originals, Russian glosses and Uzbek paraphrases are planned.
- No text of copyrighted modern editions is included; such editions are cited to verify method and answer only.
Licence
Items, gold derivations and rubric: CC BY 4.0. Wording quoted verbatim from F. Rosen, The Algebra of Mohammed ben Musa (London, 1831) is in the public domain. Harness code on GitHub: MIT.
Citation
@dataset{akhtyamov_khwarezm100_2026,
author = {Akhtyamov, Ravil},
title = {Khwarezm-100: an access-bounded benchmark of medieval Khwarezm-school mathematics},
version = {0.1.0},
year = {2026},
publisher = {Zenodo},
doi = {10.5281/zenodo.23057366}
}
@misc{akhtyamov_accessbounded_2026,
author = {Akhtyamov, Ravil},
title = {Access-Bounded Benchmarks: Evaluating LLMs Where Scaling Cannot Close the Gap (with Khwarezm-100)},
year = {2026},
doi = {10.5281/zenodo.23060373},
note = {Preprint}
}
Author: Ravil Akhtyamov (ORCID 0000-0002-0783-8573).
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